Difference between revisions of "2002 AMC 10P Problems/Problem 2"
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| − | + | == Problem 2 == | |
| + | The sum of eleven consecutive integers is <math>2002.</math> What is the smallest of these integers? | ||
| + | |||
| + | <math> | ||
| + | \text{(A) }175 | ||
| + | \qquad | ||
| + | \text{(B) }177 | ||
| + | \qquad | ||
| + | \text{(C) }179 | ||
| + | \qquad | ||
| + | \text{(D) }180 | ||
| + | \qquad | ||
| + | \text{(E) }181 | ||
| + | </math> | ||
| + | |||
| + | == Solution 1 == | ||
| + | |||
| + | == See Also == | ||
| + | {{AMC10 box|year=2002|ab=P|num-b=1|num-a=3}} | ||
| + | {{MAA Notice}} | ||
Revision as of 07:49, 15 July 2024
Problem 2
The sum of eleven consecutive integers is 
What is the smallest of these integers?
Solution 1
See Also
| 2002 AMC 10P (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
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